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If $A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&1\end{array}} \right],$ then $adj\;\left( {3{A^2} + 12A} \right) = $ . . . .
$\left[ {\begin{array}{*{20}{c}}{72}&{ - 63}\\{ - 84}&{51}\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}{72}&{ - 84}\\{ - 63}&{51}\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}{51}&{63}\\{84}&{72}\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}{51}&{84}\\{63}&{72}\end{array}} \right]$
Solution
We have $A = \left[ {\begin{array}{*{20}{c}}
2&{ – 3}\\
{ – 4}&1
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{16}&{ – 9}\\
{ – 12}&{13}
\end{array}} \right]$
$ \Rightarrow 3{A^2} = \left[ {\begin{array}{*{20}{c}}
{48}&{ – 27}\\
{ – 36}&{39}
\end{array}} \right]$
Also $12A = \left[ {\begin{array}{*{20}{c}}
{24}&{ – 36}\\
{ – 48}&{12}
\end{array}} \right]$
$\therefore 3{A^2} + 12A = \left[ {\begin{array}{*{20}{c}}
{48}&{ – 27}\\
{ – 36}&{39}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{24}&{ – 36}\\
{ – 48}&{12}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{72}&{ – 63}\\
{ – 84}&{51}
\end{array}} \right]$
adj $\left( {3{A^2} + 12A} \right) = \left[ {\begin{array}{*{20}{c}}
{51}&{63}\\
{84}&{72}
\end{array}} \right]$
Similar Questions
A manufacturer produces three products $x,\, y,\, z$ which he sells in two markets. Annual sales are indicated below:
| Market | $x$ | $y$ | $z$ |
| $I$ | $10,000$ | $2,000$ | $18,000$ |
| $II$ | $6,000$ | $20,000$ | $8,000$ |
If the unit costs of the above three commodities are $\mathrm{Rs} $. $2.00, $ $\mathrm{Rs} $. $1.00$ and $50$ paise respectively. Find the gross profit.
